Problem: The four complex roots of
\[2z^4 + 8iz^3 + (-9 + 9i)z^2 + (-18 - 2i)z + (3 - 12i) = 0,\]when plotted in the complex plane, form a rhombus.  Find the area of the rhombus.
Answer: Let $a,$ $b,$ $c,$ $d$ be the roots of the quartic.  Let $A$ be the point corresponding to complex number $a,$ etc.

Let $O$ be the center of the rhombus.  Then the complex number corresponding to $O$ is the average of $a,$ $b,$ $c,$ $d.$  By Vieta's formulas, $a + b + c + d = -\frac{8i}{2} = -4i,$ so their average is $\frac{-4i}{4} = -i.$  Hence, $O$ is located at $-i.$

[asy]
unitsize(2 cm);

pair A, B, C, D, O;

A = (-1.3362,0.8539);
C = (1.3362,-2.8539);
D = (-0.5613,-1.4046);
B = (0.5613,-0.59544);
O = (A + C)/2;

dot("$A$", A, NW);
dot("$B$", B, NE);
dot("$C$", C, SE);
dot("$D$", D, SW);
dot("$O$", O, S);

draw(A--B--C--D--cycle);
draw(A--C);
draw(B--D);

label("$p$", (A + O)/2, SW, red);
label("$q$", (B + O)/2, SE, red);
[/asy]

Let $p = OA$ and $q = OB.$  Then we want to compute the area of the rhombus, which is $4 \cdot \frac{1}{2} pq = 2pq.$

We see that $p = |a + i| = |c + i|$ and $q = |b + i| = |d + i|.$

Since $a,$ $b,$ $c,$ $d$ are the roots of the quartic in the problem, we can write
\[2z^4 + 8iz^3 + (-9 + 9i)z^2 + (-18 - 2i)z + (3 - 12i) = 2(z - a)(z - b)(z - c)(z - d).\]Setting $z = -i,$ we get
\[4 - 3i = 2(-i - a)(-i - b)(-i - c)(-i - d).\]Taking the absolute value of both sides, we get
\[5 = 2 |(a + i)(b + i)(c + i)(d + i)| = 2p^2 q^2.\]Then $4p^2 q^2 = 10,$ so $2pq = \boxed{\sqrt{10}}.$